Generalized Weighted Moore-Penrose Invertibility in Rings

1. Introduction

An involution of a ring R is an anti-automorphism whose square-root is the identity map 1. A ring R with involution * is called a *-ring. An element a in a *-ring R has group inverse provided that there exists $x\in R$ such that

$$x{a}^2=a,a{x}^2=x,ax=xa.$$

Such x is unique if exists, denoted by $a^{\#}$, and called the group inverse of a (see [1]). It is well known that a complex matrix A has group inverse if and only if $rank\left(A\right)=rank\left(A^2\right)$. Let $e\in R$ be an invertible Heimitian element (i.e., e is a unit and $e^{*}=e$). We say that $a\in R$ has e-core inverse if there exists some $x\in R$ such that

$$x{a}^2=a,a{x}^2=x,{\left(eax\right)}^{*}=eax.$$

If such x exists, it is unique, and denote it by $a_e^{\#}$ (see [2]). Many properties of e-core inverse were investigated in [3,4,5].

Let $a\in R$. Set $comm\left(a\right)=\{x\in R\mid ax=xa\}$ and $com{m}^2\left(a\right)=\{x\in R\mid xy=yx,y\in comm\left(a\right)\}.$ Recall that $a\in R$ has generalized Drazin inverse if there exists $x\in R$ such that

$$a{x}^2=x,ax=xa,a-a^{2}x\in R^{qnil}.$$

Such x is unique, if exists, and denote it by $a^d$. Here, $R^{qnil}=\{x\in R\mid 1+xy\in R^{-1}foranyy\in com{m}^2\left(x\right)\}.$ Let $e,f\in R$ be invertible Hermitian elements.

Definition 1.1.  

An element $a\in R$ has $(e,f)$-MP inverse if there exist $x\in R$ such that

$$axa=a,xax=x,{\left(eax\right)}^{*}=eax,{\left(fxa\right)}^{*}=fxa.$$

The preceding x is unique if it exists, and we denote it by $a_{(e,f)}^{†}$. The set of all $(e,f)$-MP invertible elements in R is denoted by $R_{(e,f)}^{†}$.

Recently, weighted Moore-Penrose inverses are extensively studied by many authors from very different point-views (see [3,4,6]). The motivation of this paper is to investigate when a ring element can be written as the sum of a weighted Moore-Penrose invertible element and a quasinilpotent. In Section 2, we present new characterization of the weighted Moore-Penrose inverse. It is proved that $a\in R_{(e,f)}^{†}$ if and only if ${\left(a{f}^{-1}{a}^{*}e\right)}_e^{\#}$ and $aR=\left(a{f}^{-1}{a}^{*}\right)R$. In [7] (Theorem 5), Koliha et al. expressed the weighted MP-inverse respect to a pairs of positive elements of a regular $C^{*}$-element in terms of the ordinary MP-inverse. We shall generalize this framework to encompass the wider context of ring elements.

We use $R_{(e,f)}^{qnil}$ to stand for the set $\{y\in R\mid y{f}^{-1}{y}^{*}e\in R^{qnil}\}.$ For further use, we adopt:

Definition 1.2.  

An element $a\in R$ has generalized $(e,f)$-MP inverse if there exist $x,y\in R$ such that

$$a=x+y,x^{*}ey=y{f}^{-1}{x}^{*}=0,x\in R_{(e,f)}^{†},y\in R_{(e,f)}^{qnil}.$$

We denote $x_{(e,f)}^{†}$ by $a_{(e,f)}^{†}$ and call it a generalized $(e,f)$-MP inverse of a.

In Section 3, we establish the connection between weighted Moore-Penrose inverse and generalized Drazin inverse. It is shown that $a\in R_{(e,f)}^{†}$ if and only if $a{f}^{-1}{a}^{*}e\in R^d$.

Finally, in Section 4, we further explore the generalized weighted Moore-Penrose inverse involving square-root weights. Let $\mathcal{A}$ be a *-ring and let $e,f$ be square-root elements of R. We prove that $a\in R_{e,f}^{†}$ if and only if $f^{{\displaystyle \frac{1}{2}}}a{e}^{-{\displaystyle \frac{1}{2}}}\in R^{†}$. This bridges the known formula in [7] (Theorem 5) to the broader realm of ring elements.

Throughout the paper, all *-rings are associative with an involution *. We use $R^{-1},R^{\#},R^d,R_e^{\#}$ and $R_e^{ⓓ}$ to denote the sets of all invertible, group invertible, generalized Drazin invertible and generalized e-core invertible in R, respectively. We use $\ell \left(a\right)$ to stand for the left annihilator of a.

2. Weighted Moore-Penrose Inverse

The purpose of this section is to present new characterize the weighted Moore-Penrose inverse by using the weighted core invertibility. We begin with

Lemma 2.1.  

Let $a\in R$. Then the following are equivalent:

(1)

$a\in R_{(e,f)}^{†}$.

(2)

$a\in a{f}^{-1}{a}^{*}R\bigcap R{a}^{*}ea$.

Proof. 

See [8] (Theorem 2.1). □

An element $a\in R$ is e-EP if it has e-core inverse and $a{a}_e^{\#}=a_e^{\#}a$. We now derive

Theorem 2.2.  

Let $a\in R$. Then the following are equivalent:

(1)

$a\in R_{(e,f)}^{†}$.

(2)

${\left(a{f}^{-1}{a}^{*}e\right)}^{\#}$ exists and $aR=\left(a{f}^{-1}{a}^{*}\right)R$.

(3)

${\left(a{f}^{-1}{a}^{*}e\right)}_e^{\#}$ exists and $aR=\left(a{f}^{-1}{a}^{*}\right)R$.

(4)

$a{f}^{-1}{a}^{*}e$ is e-EP and $aR=\left(a{f}^{-1}{a}^{*}\right)R$.

In this case,

$$\begin{array}{ccc}\hfill {\left(a{f}^{-1}{a}^{*}e\right)}_e^{\#}& =\hfill & e^{-1}{\left(a_{(e,f)}^{†}\right)}^{*}f{a}_{(e,f)}^{†},\hfill \\ \hfill a_{(e,f)}^{†}& =\hfill & f^{-1}{a}^{*}e{\left(a{f}^{-1}{a}^{*}e\right)}_e^{\#}.\hfill \end{array}$$

Proof.$\left(1\right)\Rightarrow \left(4\right)$ We directly check that

$$\begin{array}{ccc}\hfill \left(a{f}^{-1}{a}^{*}e\right)\left(e^{-1}{\left(a_{(e,f)}^{†}\right)}^{*}f{a}_{(e,f)}^{†}\right)& =\hfill & a{a}_{(e,f)}^{†},\hfill \\ \hfill \left(e^{-1}{\left(a_{(e,f)}^{†}\right)}^{*}f{a}_{(e,f)}^{†}\right)\left(a{f}^{-1}{a}^{*}e\right)& =\hfill & a{a}_{(e,f)}^{†}.\hfill \end{array}$$

Then ${\left(e\left(a{f}^{-1}{a}^{*}e\right)\left(e^{-1}{\left(a_{(e,f)}^{†}\right)}^{*}f{a}_{(e,f)}^{†}\right)\right)}^{*}=e\left(a{f}^{-1}{a}^{*}e\right)\left(e^{-1}{\left(a_{(e,f)}^{†}\right)}^{*}f{a}_{(e,f)}^{†}\right).$ Moreover, we have

$$\begin{array}{cc}& \left(a{f}^{-1}{a}^{*}e\right)\left(e^{-1}{\left(a_{(e,f)}^{†}\right)}^{*}f{a}_{(e,f)}^{†}\right)\left(a{f}^{-1}{a}^{*}e\right)\hfill \\ \hfill =& a{a}_{(e,f)}^{†}\left(a{f}^{-1}{a}^{*}e\right)=a{f}^{-1}{a}^{*}e,\hfill \\ & \left(e^{-1}{\left(a_{(e,f)}^{†}\right)}^{*}f{a}_{(e,f)}^{†}\right)\left(a{f}^{-1}{a}^{*}e\right)\left(e^{-1}{\left(a_{(e,f)}^{†}\right)}^{*}f{a}_{(e,f)}^{†}\right)\hfill \\ \hfill =& a{a}_{(e,f)}^{†}\left(e^{-1}{\left(a_{(e,f)}^{†}\right)}^{*}f{a}_{(e,f)}^{†}\right)=e^{-1}{\left(a_{(e,f)}^{†}\right)}^{*}f{a}_{(e,f)}^{†}.\hfill \end{array}$$

Therefore $a{f}^{-1}{a}^{*}e\in R$ is an e-EP element and ${\left(a{f}^{-1}{a}^{*}e\right)}_e^{\#}=e^{-1}{\left(a_{(e,f)}^{†}\right)}^{*}f{a}_{(e,f)}^{†}.$

In view of Lemma 2.1, we have $aR\subseteq a{f}^{-1}{a}^{*}R\subseteq aR$. Hence $aR=\left(a{f}^{-1}{a}^{*}\right)R$, as desired.

$\left(4\right)\Rightarrow \left(3\right)$ This is obvious.

$\left(3\right)\Rightarrow \left(2\right)$ By using [9] (Lemma 2.4), ${\left(a{f}^{-1}{a}^{*}e\right)}^{\#}$ exists, as desired.

$\left(2\right)\Rightarrow \left(1\right)$ This is proved in [8] (Theorem 4.4). □

Corollary 2.3.  

Let $a\in R$. Then the following are equivalent:

(1)

$a\in R_{(e,f)}^{†}$.

(2)

${\left(a{f}^{-1}{a}^{*}e\right)}^{\#}$ exists and ${⠀}^0\left(a\right){=}^0\left(a{f}^{-1}{a}^{*}\right)$.

(3)

${\left(a{f}^{-1}{a}^{*}e\right)}_e^{\#}$ and ${\left(a\right)}^0={\left(a^{*}ea\right)}^0$.

(4)

$a{f}^{-1}{a}^{*}e$ is e-EP and ${\left(a\right)}^0={\left(a^{*}ea\right)}^0$.

Proof.$\left(1\right)\Rightarrow \left(4\right)$ In view of Theorem 2.2, $a{f}^{-1}{a}^{*}e$ is e-EP and $aR=\left(a{f}^{-1}{a}^{*}e\right)R$. This implies that ${⠀}^0\left(a\right){=}^0\left(a{f}^{-1}{a}^{*}e\right)R$, as required.

$\left(4\right)\Rightarrow \left(3\right)\Rightarrow \left(2\right)$ These are obvious.

$\left(2\right)\Rightarrow \left(1\right)$ Set $z=a{f}^{-1}{a}^{*}e$ and $x=f^{-1}{a}^{*}e{z}^{\#}$. Then we see that $axz=a\left[f^{-1}{a}^{*}e{z}^{\#}\right]a{f}^{-1}{a}^{*}e=\left[a{f}^{-1}{a}^{*}e\right]z^{\#}\left[a{f}^{-1}{a}^{*}e\right]=z$, and then $(1-ax)z=0$. Hence $(1-ax)a{f}^{-1}{a}^{*}=0$. This implies that $(1-ax)a=0$; hence, $a=axa=\left(a{f}^{-1}{a}^{*}\right)e{z}^{\#}a$. Therefore $aR\subseteq \left(a{f}^{-1}{a}^{*}\right)R$, thus yielding the result by Theorem 2.2. □

Corollary 2.4.  

Let $a\in R$. Then $a\in R_{(e,f)}^{†}$ if and only if

(1)

$aR=\left(a{f}^{-1}{a}^{*}\right)R$;

(2)

there exists an idempotent p such that ${\left(ep\right)}^{*}=ep,pa{f}^{-1}{a}^{*}=0,a{f}^{-1}{a}^{*}e+p\in R^{-1}.$

In this case, $a_{(e,f)}^{†}=f^{-1}{a}^{*}e{\left((a{f}^{-1}{a}^{*}e+p\right)}^{-1}(1-p)$.

Proof. 

By virtue of Theorem 2.2, $a\in R_{(e,f)}^{†}$ if and only if $a{f}^{-1}{a}^{*}e$ has e-core inverse and $aR=\left(a{f}^{-1}{a}^{*}\right)R$. Moreover, we have

$$a_{(e,f)}^{†}=f^{-1}{a}^{*}e{\left(a{f}^{-1}{a}^{*}e\right)}^{\#}.$$

According to [] (Theorem 2.2), $a{f}^{-1}{a}^{*}e$ has e-core inverse if and only if there exists an idempotent p such that ${\left(ep\right)}^{*}=ep,pa{f}^{-1}{a}^{*}=0,a{f}^{-1}{a}^{*}e+p\in R^{-1}.$ Furthermore, we have

$${\left(a{f}^{-1}{a}^{*}e\right)}_e^{\#}={\left((a{f}^{-1}{a}^{*}e+p\right)}^{-1}(1-p),$$

thus yielding the result. □

An element a in a ring R has f-dual-core inverse if there exists some $x\in R$ such that $x{a}^2=a,a{x}^2=x,{\left(fxa\right)}^{*}=fxa$ (see [4]) Dually, we obtain the following theorem, which can be proved similarly.

Theorem 2.5.  

Let $a\in R$. Then the following are equivalent:

(1)

$a\in R_{(e,f)}^{†}$.

(2)

${\left(f^{-1}{a}^{*}ea\right)}^{\#}$ exists and $Ra=R\left(a^{*}ea\right)$.

(3)

$f^{-1}{a}^{*}ea$ has f-dual-core inverse and $Ra=R\left(a^{*}ea\right)$.

(4)

$f^{-1}{a}^{*}ea$ is f-EP and $Ra=R\left(a^{*}ea\right)$.

Corollary 2.6.  

Let $a\in R$. Then the following are equivalent:

(1)

$a\in R_{(e,f)}^{†}$.

(2)

${\left(f^{-1}{a}^{*}ea\right)}^{\#}$ exists and ${\left(a\right)}^0={\left(a^{*}ea\right)}^0$.

(3)

$f^{-1}{a}^{*}ea$ has f-dual-core inverse and ${\left(a\right)}^0={\left(a^{*}ea\right)}^0$.

(4)

$f^{-1}{a}^{*}ea$ is f-EP and ${\left(a\right)}^0={\left(a^{*}ea\right)}^0$.

We have now generalized [7] (Theorem 5) to a broader context of ring elements.

Theorem 2.7.  

Let R be a *-ring and let $e,f$ be square-root elements of R. Then the following are equivalent:

(1)

$a\in R_{e,f}^{†}$.

(2)

$e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\in R^{†}$.

In this case,

$$a_{e,f}^{†}=f^{-{\displaystyle \frac{1}{2}}}{\left(e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\right)}^{†}{e}^{{\displaystyle \frac{1}{2}}}.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Set $x=a_{e,f}^{†}$. Then $a=axa,x=xax,{\left(eax\right)}^{*}=eax,{\left(fxa\right)}^{*}=fxa$. Then we verify that

$$\begin{array}{ccc}\hfill \left(e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\right)\left(f^{{\displaystyle \frac{1}{2}}}x{e}^{-{\displaystyle \frac{1}{2}}}\right)& =\hfill & e^{{\displaystyle \frac{1}{2}}}ax{e}^{-{\displaystyle \frac{1}{2}}}=e^{-{\displaystyle \frac{1}{2}}}\left(eax\right)e^{-{\displaystyle \frac{1}{2}}},\hfill \\ \hfill \left(f^{{\displaystyle \frac{1}{2}}}x{e}^{-{\displaystyle \frac{1}{2}}}\right)\left(e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\right)& =\hfill & f^{{\displaystyle \frac{1}{2}}}xa{f}^{-{\displaystyle \frac{1}{2}}}=f^{-{\displaystyle \frac{1}{2}}}\left(fxa\right)f^{-{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

Then we have

$$\begin{array}{ccc}\hfill {\left(\left(e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\right)\left(f^{{\displaystyle \frac{1}{2}}}x{e}^{-{\displaystyle \frac{1}{2}}}\right)\right)}^{*}& =\hfill & \left(e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\right)\left(f^{{\displaystyle \frac{1}{2}}}x{e}^{-{\displaystyle \frac{1}{2}}}\right),\hfill \\ \hfill {\left(\left(f^{{\displaystyle \frac{1}{2}}}x{e}^{-{\displaystyle \frac{1}{2}}}\right)\left(e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\right)\right)}^{*}& =\hfill & \left(f^{{\displaystyle \frac{1}{2}}}x{e}^{-{\displaystyle \frac{1}{2}}}\right)\left(e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\right).\hfill \end{array}$$

Moreover, we verify that

$$\begin{array}{ccc}\hfill \left(e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\right)\left(f^{{\displaystyle \frac{1}{2}}}x{e}^{-{\displaystyle \frac{1}{2}}}\right)\left(e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\right)& =\hfill & e^{{\displaystyle \frac{1}{2}}}e\left(axa\right)f^{-{\displaystyle \frac{1}{2}}}]=e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}},\hfill \\ \hfill \left(f^{{\displaystyle \frac{1}{2}}}x{e}^{-{\displaystyle \frac{1}{2}}}\right)\left(e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\right)\left(f^{{\displaystyle \frac{1}{2}}}x{e}^{-{\displaystyle \frac{1}{2}}}\right)& =\hfill & f^{{\displaystyle \frac{1}{2}}}\left(xax\right)e^{-{\displaystyle \frac{1}{2}}}=f^{{\displaystyle \frac{1}{2}}}x{e}^{-{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

Therefore $e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\in R^{†}$ and ${\left(e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\right)}^{†}=f^{{\displaystyle \frac{1}{2}}}x{e}^{-{\displaystyle \frac{1}{2}}}.$

$\left(1\right)\Rightarrow \left(2\right)$ Set $x={\left(e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\right)}^{†}$. Then we have

$$\begin{array}{ccc}\hfill e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}x{e}^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}& =\hfill & e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}},\hfill \\ \hfill x{e}^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}x& =\hfill & x,\hfill \\ \hfill {\left(e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}x\right)}^{*}& =\hfill & e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}x,\hfill \\ \hfill {\left(x{e}^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\right)}^{*}& =\hfill & x{e}^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

We directly verify that

$$\begin{array}{ccc}\hfill a{f}^{-{\displaystyle \frac{1}{2}}}x{e}^{{\displaystyle \frac{1}{2}}}& =\hfill & e^{-{\displaystyle \frac{1}{2}}}\left(\left(e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\right)x\right)e^{{\displaystyle \frac{1}{2}}},\hfill \\ \hfill f^{-{\displaystyle \frac{1}{2}}}x{e}^{{\displaystyle \frac{1}{2}}}a& =\hfill & f^{-{\displaystyle \frac{1}{2}}}\left(x\left(e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\right)\right)f^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

Therefore

$$\begin{array}{ccc}\hfill {\left(ea{f}^{-{\displaystyle \frac{1}{2}}}x{e}^{{\displaystyle \frac{1}{2}}}\right)}^{*}& =\hfill & e^{{\displaystyle \frac{1}{2}}}\left(\left(e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\right)x\right)e^{{\displaystyle \frac{1}{2}}}=ea{f}^{-{\displaystyle \frac{1}{2}}}x{e}^{{\displaystyle \frac{1}{2}}},\hfill \\ \hfill {\left(f{f}^{-{\displaystyle \frac{1}{2}}}x{e}^{{\displaystyle \frac{1}{2}}}a\right)}^{*}& =\hfill & f^{{\displaystyle \frac{1}{2}}}\left(x\left(e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\right)\right)f^{{\displaystyle \frac{1}{2}}}=f{f}^{-{\displaystyle \frac{1}{2}}}x{e}^{{\displaystyle \frac{1}{2}}}a.\hfill \end{array}$$

Moreover, we have

$$\begin{array}{ccc}\hfill a{f}^{-{\displaystyle \frac{1}{2}}}x{e}^{{\displaystyle \frac{1}{2}}}a& =\hfill & e^{-{\displaystyle \frac{1}{2}}}\left(\left(e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\right)x\left(e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\right)\right)f^{{\displaystyle \frac{1}{2}}}\hfill \\ & =\hfill & e^{-{\displaystyle \frac{1}{2}}}\left(e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\right)f^{{\displaystyle \frac{1}{2}}}=a,\hfill \\ \hfill \left(f^{-{\displaystyle \frac{1}{2}}}x{e}^{{\displaystyle \frac{1}{2}}}\right)a\left(f^{-{\displaystyle \frac{1}{2}}}x{e}^{{\displaystyle \frac{1}{2}}}\right)& =\hfill & f^{-{\displaystyle \frac{1}{2}}}\left(x(e^{{\displaystyle \frac{1}{2}}}a\left(f^{-{\displaystyle \frac{1}{2}}}\right)x\right)e^{{\displaystyle \frac{1}{2}}}=f^{-{\displaystyle \frac{1}{2}}}x{e}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

Therefore $a_{e,f}^{†}=f^{-{\displaystyle \frac{1}{2}}}x{e}^{{\displaystyle \frac{1}{2}}},$ as required. □

Corollary 2.8.(see [7] (Theorem 5)) Let $\mathcal{A}$ be a $C^{*}$-algebra and let $e,f$ be positive. If $a\in \mathcal{A}$ be regular, then $a\in {\mathcal{A}}_{e,f}^{†}$ and

$$a_{e,f}^{†}=f^{-{\displaystyle \frac{1}{2}}}{\left(e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\right)}^{†}{e}^{{\displaystyle \frac{1}{2}}}.$$

Proof. 

Obviously, every positive elements in $\mathcal{A}$ is square-root. Since $a\in \mathcal{A}$ is regular, then $e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\in \mathcal{A}$ is regular. By virtue of [7] (Proposition 2), $e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\in {\mathcal{A}}^{†}$. Therefore we complete the proof by Theorem 2.7. □

3. Generalized Weighted MP Inverse

The purpose of this section is to introduce a new generalized inverse which is a natural generalization of group group inverse in a *-Banach algebra. Our starting points is the following.

Theorem 3.1.  

Let $a\in R$. Then the following are equivalent:

(1)

$a\in R_{(e,f)}^{†}$.

(2)

There exist $x\in R_{(e,f)}^{†},y\in R_{(e,f)}^{qnil}$ such that

$$a=x+y,x_{(e,f)}^{†}y=y{x}_{(e,f)}^{†}=0.$$

(3)

There exists $x\in R$ such that

$$x=xax,{\left(eax\right)}^{*}=eax,{\left(fxa\right)}^{*}=fxa,a-axa\in R_{(e,f)}^{qnil}.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ By hypothesis, there exist $x\in R_{(e,f)}^{†},y\in R_{(e,f)}^{qnil}$ such that

$$a=x+y,x^{*}ey=y{f}^{-1}{x}^{*}=0.$$

Then

$$\begin{array}{ccc}\hfill x_{(e,f)}^{†}y& =\hfill & x_{(e,f)}^{†}{e}^{-1}{\left(ex{x}_{(e,f)}^{†}\right)}^{*}y=x_{(e,f)}^{†}{e}^{-1}{\left(x_{(e,f)}^{†}\right)}^{*}\left(x^{*}ey\right)=0,\hfill \\ \hfill y{x}_{(e,f)}^{†}& =\hfill & y{f}^{-1}{\left(f{x}_{(e,f)}^{†}x\right)}^{*}{x}_{(e,f)}^{†}=\left(y{f}^{-1}{x}^{*}\right){\left(f{x}_{(e,f)}^{†}\right)}^{*}{x}_{(e,f)}^{†}=0,\hfill \end{array}$$

as required.

$\left(2\right)\Rightarrow \left(1\right)$ By assumption, there exist $x,y\in R$ such that

$$a=x+y,x_{(e,f)}^{†}y=y{x}_{(e,f)}^{†}=0,x\in R_{(e,f)}^{†},y\in R_{(e,f)}^{qnil}.$$

Thus, we verify that

$$\begin{array}{ccc}\hfill x^{*}ey& =\hfill & {\left(ex{x}_{(e,f)}^{†}x\right)}^{*}y=x^{*}ex\left(x_{(e,f)}^{†}y\right)=0,\hfill \\ \hfill y{f}^{-1}{x}^{*}& =\hfill & y{f}^{-1}{\left(x{x}_{(e,f)}^{†}x\right)}^{*}=y{f}^{-1}{\left(\left(x{f}^{-1}\right)\left(f{x}_{(e,f)}^{†}x\right)\right)}^{*}\hfill \\ & =\hfill & \left[y{f}^{-1}\left(f{x}_{(e,f)}^{†}x\right)\right]{\left(x{f}^{-1}\right)}^{*}=0.\hfill \end{array}$$

as desired.

$\left(1\right)\Rightarrow \left(3\right)$ By hypothesis, there exist $z,y\in R$ such that

$$a=z+y,z^{*}ey=y{f}^{-1}{z}^{*}=0,z\in R_{(e,f)}^{†},y\in R_{(e,f)}^{qnil}.$$

Set $x=z_{(e,f)}^{†}$. One easily checks that

$$\begin{array}{ccc}\hfill xy& =\hfill & z_{(e,f)}^{†}{e}^{-1}\left(ez{z}_{(e,f)}^{†}\right)y=z_{(e,f)}^{†}{e}^{-1}{\left(ez{z}_{(e,f)}^{†}\right)}^{*}y\hfill \\ & =\hfill & z_{(e,f)}^{†}{e}^{-1}{\left(z_{(e,f)}^{†}\right)}^{*}\left(z^{*}ey\right)=0,\hfill \\ \hfill xa& =\hfill & z_{(e,f)}^{†}z+z_{(e,f)}^{†}y=z_{(e,f)}^{†}z,\hfill \\ \hfill xax& =\hfill & z_{(e,f)}^{†}z{z}_{(e,f)}^{†}=z_{(e,f)}^{†}=x,\hfill \end{array}$$

Moreover, we check that

$$\begin{array}{ccc}\hfill y{z}_{(e,f)}^{†}& =\hfill & y{z}_{(e,f)}^{†}z{z}_{(e,f)}^{†}=y{f}^{-1}\left[f{z}_{(e,f)}^{†}z\right]z_{(e,f)}^{†}\hfill \\ & =\hfill & \left(y{f}^{-1}{z}^{*}\right){\left[f{z}_{(e,f)}^{†}\right)}^{*}{z}_{(e,f)}^{†}=0,\hfill \end{array}$$

and then

$$ax=(z+y)z_{(e,f)}^{†}=z{z}_{(e,f)}^{†}+y{z}_{(e,f)}^{†}=z{z}_{(e,f)}^{†}.$$

Then

$$\begin{array}{c}{\left(eax\right)}^{*}={\left(ez{z}_{(e,f)}^{†}\right)}^{*}=ez{z}_{(e,f)}^{†}=eax,\\ {\left(fxa\right)}^{*}={\left(f{z}_{(e,f)}^{†}z\right)}^{*}=f{z}_{(e,f)}^{†}z=fxa.\end{array}$$

Since $y{z}_{(e,f)}^{†}=0$, we see that

$$a-axa=a(1-xa)=(y+z)(1-z_{(e,f)}^{†}z)=y(1-z_{(e,f)}^{†}z)=y\in R_{(e,f)}^{qnil},$$

as required.

$\left(2\right)\Rightarrow \left(1\right)$ By hypotheses, there exists $z\in R$ such that

$$z=zaz,{\left(eaz\right)}^{*}=eaz,{\left(fza\right)}^{*}=fza,a-aza\in R_{(e,f)}^{qnil}.$$

Set $x=aza$ and $y=a-aza.$ Then $a=x+y$. We claim that $x\in R_{(e,f)}^{†}$. Evidently, we verify that

$$\begin{array}{ccc}\hfill xzx& =\hfill & \left(aza\right)z\left(aza\right)=a\left(zaz\right)aza=a\left(zaz\right)a=aza=x,\hfill \\ \hfill zxz& =\hfill & z\left(aza\right)z=za\left(zaz\right)=zaz=z,\hfill \\ \hfill exz& =\hfill & eazaz=eaz,\hfill \\ \hfill fzx& =\hfill & fzaza=fza,\hfill \\ \hfill {\left(exz\right)}^{*}& =\hfill & {\left(eaz\right)}^{*}=eaz=exz,\hfill \\ \hfill {\left(fzx\right)}^{*}& =\hfill & {\left(fza\right)}^{*}=fza=fzx.\hfill \end{array}$$

Therefore $x\in R_{(e,f)}^{†}$ and $z=x_{(e,f)}^{†}$. Accordingly, $x\in R_{(e,f)}^{†}.$

Moreover, we see that

$$\begin{array}{ccc}\hfill x^{*}ey& =\hfill & {\left(aza\right)}^{*}e(1-az)a=a^{*}{\left(az\right)}^{*}{e}^{*}(1-az)a\hfill \\ & =\hfill & a^{*}{\left(eaz\right)}^{*}(1-az)a=0,\hfill \\ & =\hfill & a^{*}\left(eaz\right)(1-az)a=0.\hfill \end{array}$$

Since ${\left(fza\right)}^{*}=fza$, we have $f^{-1}{\left(za\right)}^{*}{f}^{*}=za$, and so $f^{-1}{\left(za\right)}^{*}=za{f}^{-1}$. This implies that ${\left(za{f}^{-1}\right)}^{*}=za{f}^{-1}$. Therefore

$$\begin{array}{ccc}\hfill y{f}^{-1}{x}^{*}& =\hfill & (a-aza)f^{-1}{\left(aza\right)}^{*}=a(1-za)f^{-1}{\left(aza\right)}^{*}\hfill \\ & =\hfill & a{\left[(1-za)f^{-1}\right]}^{*}{\left(aza\right)}^{*}=a{\left[\left(aza\right)(1-za)f^{-1}\right]}^{*}=0.\hfill \end{array}$$

Since $a-aza\in R_{(e,f)}^{qnil}$, we see that $y\in R_{(e,f)}^{qnil}$. Therefore we have a $(e,f)$-MP decomposition $a=x+y$, as asserted. □

We denote x in Theorem 3.1 by $a_{(e,f)}^{†}$, and call it the generalized $(e,f)$-MP inverse of a.

Corollary 3.2.  

Let $a\in R$. Then the following are equivalent:

(1)

$a\in R^{†}$.

(2)

There exists a $x\in R$ such that $x=xax,{\left(ax\right)}^{*}=ax,{\left(xa\right)}^{*}=xa,(a-axa){(a-axa)}^{*}\in R^{qnil}.$

Proof. 

This is obvious by choosing $e=f=1$ in Theorem 3.1. □

Recall that a has generalized e-core invertible if there exist $x,y\in \mathcal{A}$ such that $a=x+y,x^{*}ey=yx=0,x\in {\mathcal{A}}_e^{\#},y\in {\mathcal{A}}^{qnil}.$ We use $R^{e,ⓓ}$ to denote the set of all generalized e-core invertible elements in $\mathcal{A}$. Furthermore, if $x\in \mathcal{A}$ is e-EP, we say that a is generalized e-EP. For more properties of the generalized e-core inverse, we refer the reader to [10,11]. We are ready to prove:

Theorem 3.3.  

Let $a\in R$. Then the following are equivalent:

(1)

$a\in R_{(e,f)}^{†}$.

(2)

$a{f}^{-1}{a}^{*}e\in R^d$.

(3)

$a{f}^{-1}{a}^{*}e\in R^{e,ⓓ}$.

(4)

$a{f}^{-1}{a}^{*}e$ is generalized e-EP.

Proof.$\left(1\right)\Rightarrow \left(4\right)$ By hypothesis, there exist $x,y\in R$ such that

$$a=x+y,x^{*}ey=y{f}^{-1}{x}^{*}=0,x\in R_{(e,f)}^{†},y\in R_{(e,f)}^{qnil}.$$

Set $z=x_{(e,f)}^{†}$. By virtue of Theorem 2.2, $x{f}^{-1}{x}^{*}e$ is e-EP and $xR=\left(x{f}^{-1}{x}^{*}e\right)R.$ Then we verify that

$$\begin{array}{ccc}\hfill a{f}^{-1}{a}^{*}e& =\hfill & (x+y)f^{-1}(x^{*}+y^{*})e=(x+y)f^{-1}(x^{*}e+y^{*}e)\hfill \\ & =\hfill & x{f}^{-1}(x^{*}e+y^{*}e)+y{f}^{-1}(x^{*}e+y^{*}e)\hfill \\ & =\hfill & x{f}^{-1}{x}^{*}e+y{f}^{-1}{y}^{*}e.\hfill \end{array}$$

Obviously, we have

$${\left[x{f}^{-1}{x}^{*}e\right]}^{*}e\left[y{f}^{-1}{y}^{*}e\right]={\left[f^{-1}{x}^{*}e\right]}^{*}\left[x^{*}ey\right]f^{-1}{y}^{*}e=0.$$

$$\left[y{f}^{-1}{y}^{*}e\right]\left[x{f}^{-1}{x}^{*}e\right]=y{f}^{-1}\left[y^{*}ex\right]f^{-1}{x}^{*}e=0.$$

Since $x{f}^{-1}{x}^{*}e\in R$ is e-EP and $y{f}^{-1}{y}^{*}e\in R^{qnil}$. Thus $a{f}^{-1}{a}^{*}e\in R$ is generalized e-EP, as required.

$\left(4\right)\Rightarrow \left(3\right)$ This is obvious.

$\left(3\right)\Rightarrow \left(2\right)$ This is obvious by [10] (Theorem 2.5).

$\left(2\right)\Rightarrow \left(1\right)$ Let $s=a{f}^{-1}{a}^{*}e$. Then $s^{*}=ea{f}^{-1}{a}^{*}$. Hence, ${\left(s^d\right)}^{*}={\left(s^{*}\right)}^d={\left[ea{f}^{-1}{a}^{*}\right]}^d=e{\left({\left[a{f}^{-1}{a}^{*}e\right]}^d\right)}^{2}a{f}^{-1}{a}^{*}=e{\left(s^d\right)}^{2}a{f}^{-1}{a}^{*}$.

$${(1-s{s}^d)}^{*}=1-e{\left(s^d\right)}^2\left[a{f}^{-1}{a}^{*}e\right]a{f}^{-1}{a}^{*}=1-e{s}^{d}a{f}^{-1}{a}^{*}$$

$$e^{-1}{(1-s{s}^d)}^{*}e=e^{-1}[1-e{s}^{d}a{f}^{-1}{a}^{*}]e=1-s^{d}a{f}^{-1}{a}^{*}e=1-s{s}^d$$

Let $x=s{s}^{d}a$ and $y=(1-s{s}^d)a$. Then $a=x+y$. Then $x^{*}=a^{*}{s}^{*}{\left(s^d\right)}^{*}=a^{*}e\left[a{f}^{-1}{a}^{*}e\right]{\left(s^d\right)}^{2}a{f}^{-1}{a}^{*}=a^{*}e{s}^{d}a{f}^{-1}{a}^{*}$.

$$\begin{array}{ccc}\hfill x^{*}ey& =\hfill & a^{*}e{s}^{d}a{f}^{-1}{a}^{*}e(1-s{s}^d)a=a^{*}e{s}^d\left[a{f}^{-1}{a}^{*}e\right](1-s{s}^d)a\hfill \\ & =\hfill & a^{*}e{s}^{d}s(1-s{s}^d)a=0,\hfill \\ \hfill y{f}^{-1}{x}^{*}& =\hfill & (1-s{s}^d)a{f}^{-1}{a}^{*}e{s}^{d}a{f}^{-1}{a}^{*}\hfill \\ & =\hfill & (1-s{s}^d)s{s}^{d}a{f}^{-1}{a}^{*}=0.\hfill \end{array}$$

$$\begin{array}{ccc}\hfill y{f}^{-1}{y}^{*}e& =\hfill & (1-s{s}^d)a{f}^{-1}{a}^{*}{(1-s{s}^d)}^{*}\hfill \\ & =\hfill & (1-s{s}^d)\left[a{f}^{-1}{a}^{*}e\right]e^{-1}{(1-s{s}^d)}^{*}e\hfill \\ & =\hfill & (s-s^2{s}^d)\left[e^{-1}{(1-s{s}^d)}^{*}e\right]\hfill \\ & =\hfill & (s-s^2{s}^d)(1-s{s}^d)=s-s^2{s}^d\hfill \\ \hfill .\end{array}$$

Therefore we have $y{f}^{-1}{y}^{*}e\in R^{qnil}$. That is, $y\in R_{(e,f)}^{qnil}$.

We will suffice to prove that $x=s{s}^{d}a\in R_{(e,f)}^{†}$.

One directly verify that

$$\begin{array}{ccc}\hfill x{f}^{-1}{x}^{*}e& =\hfill & s{s}^{d}a{f}^{-1}{a}^{*}{\left(s{s}^d\right)}^{*}e\hfill \\ & =\hfill & s{s}^d\left[a{f}^{-1}{a}^{*}e\right]e^{-1}{\left(s{s}^d\right)}^{*}e\hfill \\ & =\hfill & s{s}^{d}s\left[e^{-1}{\left(s{s}^d\right)}^{*}e\right]\hfill \\ & =\hfill & s{s}^{d}s\left(s{s}^d\right)=s^2{s}^d.\hfill \end{array}$$

Hence, $x{f}^{-1}{x}^{*}e\in R^{\#}$ and ${\left(x{f}^{-1}{x}^{*}e\right)}^{\#}=s^d$.

On the other hand, we have

$$xR=s{s}^{d}aR=\left(s^2{s}^d\right)\left(s^{d}a\right)R\subseteq s^2{s}^{d}R=x{f}^{-1}{x}^{*}eR\subseteq xR.$$

Thus, $xR=x{f}^{-1}{x}^{*}eR.$ By virtue of Theorem 2.2, $x=s{s}^{d}a\in R_{(e,f)}^{†}$, as required. □

Corollary 3.4.  

Let $a,b\in R_{(e,f)}^{†}$. If $a^{*}eb=b{f}^{-1}{a}^{*}=0$, then $a+b\in R_{(e,f)}^{†}$.

Proof. 

By virtue of Theorem 3.3, $a{f}^{-1}{a}^{*}e,b{f}^{-1}{b}^{*}e\in R^d$. Since $a^{*}eb=b{f}^{-1}{a}^{*}=0$, we easily check that $(a+b)f^{-1}{(a+b)}^{*}e=a{f}^{-1}{a}^{*}e+b{f}^{-1}{b}^{*}e$. Obviously, $\left[a{f}^{-1}{a}^{*}e\right]\left[b{f}^{-1}{b}^{*}e\right]=0$. According to [12](Lemma 15.2.2.], $(a+b)f^{-1}{(a+b)}^{*}e\in R^d$. Applying Theorem 3.3 again, $a+b\in R_{(e,f)}^{†}$. □

Corollary 3.5.  

Let $a\in R$. Then the following are equivalent:

(1)

$a\in R^{†}$.

(2)

There exist $x\in R^{†},y\in R$ such that

$$a=x+y,x^{†}y=y{x}^{†}=0,y{y}^{*}\in R^{qnil}.$$

(3)

$a{a}^{*}\in R^d$.

(4)

$a{a}^{*}\in R^{ⓓ}$.

(5)

$a{a}^{*}\in R^{ⓔ}$.

(6)

There exists an idempotent $p\in com{m}^2\left(a{a}^{*}\right)$ such that

$$a{a}^{*}+p\in R^{-1},a{a}^{*}p\in R^{qnil}.$$

Proof. 

This is proved by choosing $e=f=1$ in Theorem 3.3. □

The involution * is proper, that is, $x^{*}x=0⟹x=0$, e.g., in a Rickart *-ring, the involution is always proper. Let $C^{n\times n}$ be the Banach algebra of all $n\times n$ complex matrices, with conjugate transpose * as the involution. Then the involution * is proper. In [13], Zou et al. extend the notion of weak group inverse to elements in a ring with proper involution. We refer the reader for weak group inverse in [14,15,16].

Corollary 3.6.  

Let R be a ring with proper involution and $a\in R$. Then the following are equivalent:

(1)

$a\in R^{†}$.

(2)

$a{a}^{*}\in R^{ⓖ}$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ By hypothesis, there exist $x,y\in R$ such that

$$a=x+y,x^{*}y=y{x}^{*}=0,x\in R^{†},y\in R^{qnil}.$$

Set $z=x^{†}$. By virtue of Theorem 2.2, ${\left(x{x}^{*}\right)}^{\#}$ exists and $xR=\left(x{x}^{*}\right)R.$ Then we have $a{a}^{*}=x{x}^{*}+y{y}^{*},{\left[x{x}^{*}\right]}^{*}\left[y{y}^{*}\right]=x\left[x^{*}y\right]y^{*}=0$ and $\left[y{y}^{*}\right]\left[x{x}^{*}\right]=y\left[y^{*}x\right]x^{*}=0.$ Since $x{x}^{*}\in R^{\#}$ and $y{y}^{*}\in R^{qnil}$. By virtue of [17](Theorem 1.1), $a{a}^{*}\in R^{ⓖ}$.

$\left(2\right)\Rightarrow \left(1\right)$ By virtue of [17] (Theorem 1.1), $a{a}^{*}\in R^d$. The proof is completed by Corollary 3.5. □

An element a in a ring R has a generalized f-dual-core inverse if there exist $x,y\in \mathcal{A}$ such that $a=x+y,x^{*}fy=yx=0$, $x\in \mathcal{A}$ has f-dual-core inverse, $y\in {\mathcal{A}}^{qnil}.$ Dually, applying a similar argument, we deduce that

Theorem 3.7.  

Let $a\in R$. Then the following are equivalent:

(1)

$a\in R_{(e,f)}^{†}$.

(2)

$f^{-1}{a}^{*}ea\in R^d$.

(3)

$f^{-1}{a}^{*}ea\in R$ is generalized f-dual-core inverse.

(4)

$f^{-1}{a}^{*}ea$ is generalized e-EP.

The following example illustrates that the existence of a generalized MP invertibility is not preserved under the generalized Drazin invertibility, even in a $C^{*}$-algebraic setting.

Example 3.8.

Let $\mathcal{H}=\mathcal{B}\left({\ell }^2\left(N\right)\right)$. Define the operator $\alpha \in \mathcal{H}$ as follows:

$$\alpha e_n=\{\begin{array}{ccc}{\displaystyle \frac{1}{n}}{e}_{n+1},\hfill & & nisodd,\hfill \\ 0,\hfill & & niseven.\hfill \end{array}$$

Here, $\left\{e_n\right\}$ is a basis of ${\ell }^2\left(N\right)$. Then the adjoint operator ${\alpha }^{*}$ is given by

$${\alpha }^{*}{e}_n=\{\begin{array}{ccc}{\displaystyle \frac{1}{n-1}}{e}_{n-1},\hfill & & niseven,\hfill \\ 0,\hfill & & nisodd.\hfill \end{array}$$

Thus the product $\alpha {\alpha }^{*}$ is given by

$$\alpha {\alpha }^{*}{e}_n=\{\begin{array}{ccc}{\displaystyle \frac{1}{n^2}}{e}_n,\hfill & & niseven,\hfill \\ 0,\hfill & & nisodd.\hfill \end{array}$$

Obviously, $\alpha$ has a generalized Drazin inverse, as ${\alpha }^2$ is the zero operator. We claim that the element $\alpha {\alpha }^{*}$ does not possess a generalized Drazin inverse. The spectrum of $\alpha {\alpha }^{*}$ is $\sigma \left(\alpha {\alpha }^{*}\right)=\left\{0\right\}\bigcap \{{\displaystyle \frac{1}{n^2}}:n\in N\}$. Since $\underset{n\to \infty }{lim}{\displaystyle \frac{1}{n^2}}=0$, the point 0 is a limit point (or accumulation point) of the spectrum. Then 0 is not an isolated point of $\sigma \left(\alpha {\alpha }^{*}\right)$. This implies that $\alpha {\alpha }^{*}$ has no generalized Drazin inverse, because a necessary condition for an element to be generalized Drazin invertible is that 0 must be an isolated point of its spectrum. By virtue of Corollary 3.5, $\alpha$ has no generalized MP inverse.

4. Generalized MP Inverses Associated with Square-Root Weights

In this section, we are concerned with the generalized weighted Moore-Penrose inverse involving square-root weights. We shall characterize the generalized weighted Moore-Penrose inverse by using the ordinary Moore-Penrose inverse.

Lemma 4.1.  

Let $a\in R_{(e,f)}^{†},b\in R_{(e,f)}^{†}$. If $a^{*}eb=b{f}^{-1}{a}^{*}=0$, then

$${(a+b)}_{(e,f)}^{†}=a_{(e,f)}^{†}+b_{(e,f)}^{†}.$$

Proof. 

Step 1. $a,b\in R_{(e,f)}^{†}$. By hypothesis, we have $a{b}_{(e,f)}^{†}=b_{(e,f)}^{†}a=b{a}_{(e,f)}^{†}=a_{(e,f)}^{†}b=0$. Then we directly check that $a+b\in R_{(e,f)}^{†}$ and

$${(a+b)}_{(e,f)}^{†}={(a+b)}_{(e,f)}^{†}=a_{(e,f)}^{†}+b_{(e,f)}^{†}.$$

Step 2. $a\in R_{(e,f)}^{†},b\in R_{(e,f)}^{†}$. Then we can find $x\in R_{(e,f)}^{†}$ and $y\in R_{(e,f)}^{qnil}$ such that

$$a=x+y,x^{*}ey=y{f}^{-1}{x}^{*}=0.$$

As in the proof of Theorem 3.1, $x=a{a}_{(e,f)}^{†}a$ and $y=a-a{a}_{(e,f)}^{†}a$. Then $a+b=(x+b)+y$. Moreover, we see that

$$x^{*}eb={\left(a_{(e,f)}^{†}a\right)}^{*}\left(a^{*}eb\right)=0,b{f}^{-1}{x}^{*}=\left(b{f}^{-1}{a}^{*}\right){\left(a{a}_{(e,f)}^{†}\right)}^{*}=0.$$

By using Step 1, we prove that $x+b\in R_{(e,f)}^{†}$. It is easy to verify that

$$\begin{array}{ccc}\hfill {(x+b)}^{*}ey& =\hfill & x^{*}ey+b^{*}ea(1-a_{(e,f)}^{†}a)=0,\hfill \\ \hfill y{f}^{-1}{(x+b)}^{*}& =\hfill & y{f}^{-1}{x}^{*}+(1-a{a}_{(e,f)}^{†})\left(a{f}^{-1}{b}^{*}\right)=0.\hfill \end{array}$$

Therefore

$${(a+b)}_{(e,f)}^{†}={(x+b)}_{(e,f)}^{†}=x_{(e,f)}^{†}+b_{(e,f)}^{†}=a_{(e,f)}^{†}+b_{(e,f)}^{†}.$$

□

We are ready to prove:

Theorem 4.2.  

Let $p\in R$ be a projection and $x\in R$. Let $\alpha =e^{-{\displaystyle \frac{1}{2}}}pxp{f}^{{\displaystyle \frac{1}{2}}}$ and $\beta =e^{-{\displaystyle \frac{1}{2}}}pxp{f}^{{\displaystyle \frac{1}{2}}}+e^{-{\displaystyle \frac{1}{2}}}(1-p)f^{{\displaystyle \frac{1}{2}}}$. Then $\alpha \in R_{(e,f)}^{†}$ if and only if

(1)

$\beta \in R_{(e,f)}^{†};$

(2)

$\left[p{f}^{{\displaystyle \frac{1}{2}}}\right]{\beta }_{(e,f)}^{†}\left[e^{-{\displaystyle \frac{1}{2}}}(1-p)\right]=[1-\beta {\beta }_{(e,f)}^{†}]\left[e^{-{\displaystyle \frac{1}{2}}}(1-p)\right]=0$.

In this case,

$$\begin{array}{ccc}\hfill {\beta }_{(e,f)}^{†}& =\hfill & {\alpha }_{(e,f)}^{†}+f^{-{\displaystyle \frac{1}{2}}}(1-p)e^{{\displaystyle \frac{1}{2}}},\hfill \\ \hfill {\alpha }_{(e,f)}^{†}& =\hfill & f^{-{\displaystyle \frac{1}{2}}}p{f}^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}{e}^{-{\displaystyle \frac{1}{2}}}p{e}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

Proof. 

⟹ Let $\gamma =e^{-{\displaystyle \frac{1}{2}}}(1-p)f^{{\displaystyle \frac{1}{2}}}$. By virtue of Theorem 2.7, $\gamma \in R_{(e,f)}^{†}$ and ${\gamma }_{(e,f)}^{†}=f^{-{\displaystyle \frac{1}{2}}}(1-p)e^{{\displaystyle \frac{1}{2}}}$.

It is easy to verify that

$${\alpha }^{*}e\gamma =0,\gamma f^{-1}\alpha =0.$$

By hypothesis, $\alpha \in R_{(e,f)}^{†}$. In light of Lemma 4.1, $\beta \in R_{(e,f)}^{†}$ and

$$\begin{array}{ccc}\hfill {\beta }_{(e,f)}^{†}& =\hfill & {\alpha }_{(e,f)}^{†}+{\gamma }_{(e,f)}^{†}\hfill \\ & =\hfill & {\alpha }_{(e,f)}^{†}+f^{-{\displaystyle \frac{1}{2}}}(1-p)e^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

We easily check that

$$\begin{array}{cc}& \left[p{f}^{{\displaystyle \frac{1}{2}}}\right]{\beta }_{(e,f)}^{†}\left[e^{-{\displaystyle \frac{1}{2}}}(1-p)\right]\hfill \\ \hfill =& \left[p{f}^{{\displaystyle \frac{1}{2}}}\right][{\alpha }_{(e,f)}^{†}+f^{-{\displaystyle \frac{1}{2}}}(1-p)e^{{\displaystyle \frac{1}{2}}}]\left[e^{-{\displaystyle \frac{1}{2}}}(1-p)\right]\hfill \\ \hfill =& \left[p{f}^{{\displaystyle \frac{1}{2}}}\right]{\alpha }_{(e,f)}^{†}\left[e^{-{\displaystyle \frac{1}{2}}}(1-p)\right]\hfill \\ \hfill =& \left[p{f}^{{\displaystyle \frac{1}{2}}}\right]{\alpha }_{(e,f)}^{†}{e}^{-1}\left[e\alpha {\alpha }_{(e,f)}^{†}\right]\left[e^{-{\displaystyle \frac{1}{2}}}(1-p)\right]\hfill \\ \hfill =& \left[p{f}^{{\displaystyle \frac{1}{2}}}\right]{\alpha }_{(e,f)}^{†}{e}^{-1}{\left[e\alpha {\alpha }_{(e,f)}^{†}\right]}^{*}\left[e^{-{\displaystyle \frac{1}{2}}}(1-p)\right]\hfill \\ \hfill =& \left[p{f}^{{\displaystyle \frac{1}{2}}}\right]{\alpha }_{(e,f)}^{†}{e}^{-1}{\left[{\alpha }_{(e,f)}^{†}\right]}^{*}{\alpha }^{*}\left[e{e}^{-{\displaystyle \frac{1}{2}}}(1-p)\right]\hfill \\ \hfill =& \left[p{f}^{{\displaystyle \frac{1}{2}}}\right]{\alpha }_{(e,f)}^{†}{e}^{-1}{\left[{\alpha }_{(e,f)}^{†}\right]}^{*}{\left[e^{-{\displaystyle \frac{1}{2}}}pxp{f}^{{\displaystyle \frac{1}{2}}}\right]}^{*}\left[e^{{\displaystyle \frac{1}{2}}}(1-p)\right]\hfill \\ \hfill =& \left[p{f}^{{\displaystyle \frac{1}{2}}}\right]{\alpha }_{(e,f)}^{†}{e}^{-1}{\left[{\alpha }_{(e,f)}^{†}\right]}^{*}{\left[xp{f}^{{\displaystyle \frac{1}{2}}}\right]}^{*}p{e}^{-{\displaystyle \frac{1}{2}}}\left[e^{{\displaystyle \frac{1}{2}}}(1-p)\right]\hfill \\ \hfill =& 0.\hfill \end{array}$$

Moreover, we have $\beta {\beta }_{(e,f)}^{†}{e}^{-{\displaystyle \frac{1}{2}}}(1-p)f^{{\displaystyle \frac{1}{2}}}=e^{-{\displaystyle \frac{1}{2}}}(1-p)f^{{\displaystyle \frac{1}{2}}},$ and then

$$(1-\beta {\beta }_{(e,f)}^{†})e^{-{\displaystyle \frac{1}{2}}}(1-p)f^{{\displaystyle \frac{1}{2}}}=0,$$

as required.

⟸ Let $z=f^{-{\displaystyle \frac{1}{2}}}p{f}^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}{e}^{-{\displaystyle \frac{1}{2}}}p{e}^{{\displaystyle \frac{1}{2}}}$. Then we check that

$$\begin{array}{ccc}\hfill \alpha z& =\hfill & e^{-{\displaystyle \frac{1}{2}}}pxp{f}^{{\displaystyle \frac{1}{2}}}\left[f^{-{\displaystyle \frac{1}{2}}}p{f}^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}{e}^{-{\displaystyle \frac{1}{2}}}p{e}^{{\displaystyle \frac{1}{2}}}\right]\hfill \\ & =\hfill & e^{-{\displaystyle \frac{1}{2}}}pxp{f}^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}{e}^{-{\displaystyle \frac{1}{2}}}p{e}^{{\displaystyle \frac{1}{2}}},\hfill \end{array}$$

and so

$$\begin{array}{ccc}\hfill z\alpha z& =\hfill & \left[f^{-{\displaystyle \frac{1}{2}}}p{f}^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}{e}^{-{\displaystyle \frac{1}{2}}}p{e}^{{\displaystyle \frac{1}{2}}}\right]\left[e^{-{\displaystyle \frac{1}{2}}}pxp{f}^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}{e}^{-{\displaystyle \frac{1}{2}}}p{e}^{{\displaystyle \frac{1}{2}}}\right]\hfill \\ & =\hfill & f^{-{\displaystyle \frac{1}{2}}}p{f}^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}{e}^{-{\displaystyle \frac{1}{2}}}pxp{f}^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}{e}^{-{\displaystyle \frac{1}{2}}}p{e}^{{\displaystyle \frac{1}{2}}}\hfill \\ & =\hfill & f^{-{\displaystyle \frac{1}{2}}}p{f}^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}[e^{-{\displaystyle \frac{1}{2}}}pxp{f}^{{\displaystyle \frac{1}{2}}}+e^{-{\displaystyle \frac{1}{2}}}(1-p)f^{{\displaystyle \frac{1}{2}}}]{\beta }_{(e,f)}^{†}{e}^{-{\displaystyle \frac{1}{2}}}p{e}^{{\displaystyle \frac{1}{2}}}\hfill \\ & =\hfill & f^{-{\displaystyle \frac{1}{2}}}p{f}^{{\displaystyle \frac{1}{2}}}\left[{\beta }_{(e,f)}^{†}\beta {\beta }_{(e,f)}^{†}\right]e^{-{\displaystyle \frac{1}{2}}}p{e}^{{\displaystyle \frac{1}{2}}}\hfill \\ & =\hfill & f^{-{\displaystyle \frac{1}{2}}}p{f}^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}{e}^{-{\displaystyle \frac{1}{2}}}p{e}^{{\displaystyle \frac{1}{2}}}\hfill \\ & =\hfill & z.\hfill \end{array}$$

$$\begin{array}{cc}& \alpha z\alpha -\alpha \hfill \\ \hfill =& e^{-{\displaystyle \frac{1}{2}}}pxp{f}^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}{e}^{-{\displaystyle \frac{1}{2}}}pxp{f}^{{\displaystyle \frac{1}{2}}}-e^{-{\displaystyle \frac{1}{2}}}pxp{f}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill =& [e^{-{\displaystyle \frac{1}{2}}}pxp{f}^{{\displaystyle \frac{1}{2}}}+e^{-{\displaystyle \frac{1}{2}}}(1-p)f^{{\displaystyle \frac{1}{2}}}]{\beta }_{(e,f)}^{†}\left[e^{-{\displaystyle \frac{1}{2}}}pxp{f}^{{\displaystyle \frac{1}{2}}}\right]-e^{-{\displaystyle \frac{1}{2}}}pxp{f}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill =& \beta {\beta }_{(e,f)}^{†}\left[e^{-{\displaystyle \frac{1}{2}}}pxp{f}^{{\displaystyle \frac{1}{2}}}\right]-e^{-{\displaystyle \frac{1}{2}}}pxp{f}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill =& \beta {\beta }_{(e,f)}^{†}[e^{-{\displaystyle \frac{1}{2}}}pxp{f}^{{\displaystyle \frac{1}{2}}}+e^{-{\displaystyle \frac{1}{2}}}(1-p)f^{{\displaystyle \frac{1}{2}}}]-[e^{-{\displaystyle \frac{1}{2}}}(1-p)f^{{\displaystyle \frac{1}{2}}}+e^{-{\displaystyle \frac{1}{2}}}pxp{f}^{{\displaystyle \frac{1}{2}}}]\hfill \\ \hfill =& \beta {\beta }_{(e,f)}^{†}\beta -\beta .\hfill \end{array}$$

Hence,

$$\alpha z\alpha -\alpha =\beta {\beta }_{(e,f)}^{†}\beta -\beta \in R_{(e,f)}^{qnil}.$$

Since ${\left[e\beta {\beta }_{(e,f)}^{†}\right]}^{*}=e\beta {\beta }_{(e,f)}^{†},$ we have

$$\begin{array}{cc}& {\left[e(e^{-{\displaystyle \frac{1}{2}}}pxp{f}^{{\displaystyle \frac{1}{2}}}+e^{-{\displaystyle \frac{1}{2}}}(1-p)f^{{\displaystyle \frac{1}{2}}}){\beta }_{(e,f)}^{†}\right]}^{*}\hfill \\ \hfill =& e(e^{-{\displaystyle \frac{1}{2}}}pxp{f}^{{\displaystyle \frac{1}{2}}}+e^{-{\displaystyle \frac{1}{2}}}(1-p)f^{{\displaystyle \frac{1}{2}}}){\beta }_{(e,f)}^{†}.\hfill \end{array}$$

Then

$${\left[e^{{\displaystyle \frac{1}{2}}}(pxp+1-p)f^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}\right]}^{*}=e^{{\displaystyle \frac{1}{2}}}(pxp+1-p)f^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}.$$

This implies that

$${\left[(pxp+1-p)f^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}{e}^{-{\displaystyle \frac{1}{2}}}\right]}^{*}=(pxp+1-p)f^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}{e}^{-{\displaystyle \frac{1}{2}}}.$$

Hence,

$${\left[\left(pxp\right)f^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}{e}^{-{\displaystyle \frac{1}{2}}}p\right]}^{*}=\left(pxp\right)f^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}{e}^{-{\displaystyle \frac{1}{2}}}p.$$

Therefore

$$\begin{array}{cc}& {\left[e\left(e^{-{\displaystyle \frac{1}{2}}}\left(pxp\right)f^{{\displaystyle \frac{1}{2}}}\right)\left(f^{-{\displaystyle \frac{1}{2}}}p{f}^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}{e}^{-{\displaystyle \frac{1}{2}}}p{e}^{{\displaystyle \frac{1}{2}}}\right)\right]}^{*}\hfill \\ \hfill =& e\left(e^{-{\displaystyle \frac{1}{2}}}\left(pxp\right)f^{{\displaystyle \frac{1}{2}}}\right)\left(f^{-{\displaystyle \frac{1}{2}}}p{f}^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}\right)e^{-{\displaystyle \frac{1}{2}}}p{e}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

That is, ${\left(e\alpha z\right)}^{*}=e\alpha z.$ Likewise, we verify that ${\left(fz\alpha \right)}^{*}=fz\alpha$. Accordingly, $\alpha \in R_{(e,f)}^{†}$. In this case, we have

$${\alpha }_{(e,f)}^{†}=z=f^{-{\displaystyle \frac{1}{2}}}p{f}^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}{e}^{-{\displaystyle \frac{1}{2}}}p{e}^{{\displaystyle \frac{1}{2}}},$$

as asserted. □

Corollary 4.3.  

Let $p\in R$ be a projection and $x\in R$. Let $\alpha =pxp$ and $\beta =pxp+1-p$. Then $\alpha \in R^{†}$ if and only if

(1)

$\beta \in R^{†};$

(2)

$p{\beta }^{†}(1-p)=(1-\beta {\beta }^{†})(1-p)=0$.

In this case,

$$\begin{array}{ccc}\hfill {\beta }^{†}& =\hfill & {\alpha }^{†}+1-p,\hfill \\ \hfill {\alpha }^{†}& =\hfill & f^{-{\displaystyle \frac{1}{2}}}p{f}^{{\displaystyle \frac{1}{2}}}{\beta }^{†}p.\hfill \end{array}$$

Proof. 

This is obvious by choosing $e=f=1$ in Theorem 4.2. □

Theorem 4.4.  

Let $a\in R_{(e,f)}^{†},p:=a{a}_{(e,f)}^{†}$ be a projection and $x\in R$. Let $\gamma =f^{-{\displaystyle \frac{1}{2}}}\left(a_{(e,f)}^{†}xa\right)e^{{\displaystyle \frac{1}{2}}}$ and $\delta =f^{-{\displaystyle \frac{1}{2}}}\left(a_{(e,f)}^{†}xa\right)e^{{\displaystyle \frac{1}{2}}}+f^{-{\displaystyle \frac{1}{2}}}(1-a_{(e,f)}^{†}a)e^{{\displaystyle \frac{1}{2}}}$. Then $\gamma \in R_{(e,f)}^{†}$ if and only if

(1)

$\delta \in R_{(e,f)}^{†};$

(2)

$\begin{array}{c}\hfill e^{{\displaystyle \frac{1}{2}}}{\delta }_{(e,f)}^{†}\left(f^{-{\displaystyle \frac{1}{2}}}(1-a_{(e,f)}^{†}a)\right)=1-a_{(e,f)}^{†}a,\\ \hfill \left((1-a_{(e,f)}^{†}a)e^{{\displaystyle \frac{1}{2}}}\right){\delta }_{(e,f)}^{†}\left(f^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}\right)=0.\end{array}$

In this case,

$$\begin{array}{ccc}\hfill {\delta }_{(e,f)}^{†}& =\hfill & {\alpha }_{(e,f)}^{†}+e^{-{\displaystyle \frac{1}{2}}}(1-a{a}_{(e,f)}^{†})f^{{\displaystyle \frac{1}{2}}},\hfill \\ \hfill {\gamma }_{(e,f)}^{†}& =\hfill & e^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}a{e}^{{\displaystyle \frac{1}{2}}}{\delta }_{(e,f)}^{†}{f}^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}a{f}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

Proof. 

⟹ Let $\eta =f^{-{\displaystyle \frac{1}{2}}}(1-a_{(e,f)}^{†}a)e^{{\displaystyle \frac{1}{2}}}$. By virtue of Theorem 2.7, $\eta \in R_{(e,f)}^{†}$ and ${\eta }_{(e,f)}^{†}=e^{-{\displaystyle \frac{1}{2}}}(1-a_{(e,f)}^{†}a)f^{{\displaystyle \frac{1}{2}}}$.

It is easy to verify that

$$\begin{array}{ccc}\hfill {\gamma }^{*}f\eta & =\hfill & {\left[a_{(e,f)}^{†}xa{e}^{{\displaystyle \frac{1}{2}}}\right]}^{*}{f}^{-{\displaystyle \frac{1}{2}}}f\left[f^{-{\displaystyle \frac{1}{2}}}(1-a_{(e,f)}^{†}a)e^{{\displaystyle \frac{1}{2}}}\right]\hfill \\ & =\hfill & {\left[a_{(e,f)}^{†}xa{e}^{{\displaystyle \frac{1}{2}}}\right]}^{*}(1-a_{(e,f)}^{†}a)e^{{\displaystyle \frac{1}{2}}}=0,\hfill \\ \hfill \eta e^{-1}{\gamma }^{*}& =\hfill & f^{-{\displaystyle \frac{1}{2}}}(1-a_{(e,f)}^{†}a)e^{{\displaystyle \frac{1}{2}}}{e}^{-1}{\left[f^{-{\displaystyle \frac{1}{2}}}\left(a_{(e,f)}^{†}xa\right)e^{{\displaystyle \frac{1}{2}}}\right]}^{*}=0.\hfill \end{array}$$

By hypothesis, $\gamma \in R_{(f,e)}^{†}$. By virtue of Lemma 4.1, $\delta \in R_{(e,f)}^{†}$ and

$$\begin{array}{ccc}\hfill {\delta }_{(e,f)}^{†}& =\hfill & {\gamma }_{(e,f)}^{†}+{\eta }_{(e,f)}^{†}\hfill \\ & =\hfill & {\gamma }_{(e,f)}^{†}+e^{-{\displaystyle \frac{1}{2}}}(1-a_{(e,f)}^{†}a)f^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

We easily check that

$$\begin{array}{ccc}\hfill e^{{\displaystyle \frac{1}{2}}}{\delta }_{(e,f)}^{†}\left(f^{-{\displaystyle \frac{1}{2}}}(1-a_{(e,f)}^{†}a)\right)& =\hfill & 1-a_{(e,f)}^{†}a,\hfill \\ \hfill \left((1-a_{(e,f)}^{†}a)e^{{\displaystyle \frac{1}{2}}}\right]){\delta }_{(e,f)}^{†}\left(f^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}\right)& =\hfill & 0.\hfill \end{array}$$

⟸ Let $z=e^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}a{e}^{{\displaystyle \frac{1}{2}}}{\delta }_{(e,f)}^{†}{f}^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}a{f}^{{\displaystyle \frac{1}{2}}}$. Then we check that

$$\begin{array}{ccc}\hfill \gamma z& =\hfill & \left[f^{-{\displaystyle \frac{1}{2}}}\left(a_{(e,f)}^{†}xa\right)e^{{\displaystyle \frac{1}{2}}}\right]\left[e^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}a{e}^{{\displaystyle \frac{1}{2}}}{\delta }_{(e,f)}^{†}{f}^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}a{f}^{{\displaystyle \frac{1}{2}}}\right]\hfill \\ & =\hfill & f^{-{\displaystyle \frac{1}{2}}}(a_{(e,f)}^{†}xa{e}^{{\displaystyle \frac{1}{2}}}{\delta }_{(e,f)}^{†}{f}^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}a{f}^{{\displaystyle \frac{1}{2}}},\hfill \end{array}$$

and then

$$\begin{array}{cc}& z\gamma z\hfill \\ \hfill =& \left[e^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}a{e}^{{\displaystyle \frac{1}{2}}}{\delta }_{(e,f)}^{†}{f}^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}a{f}^{{\displaystyle \frac{1}{2}}}\right][f^{-{\displaystyle \frac{1}{2}}}\left(a_{(e,f)}^{†}xa{e}^{{\displaystyle \frac{1}{2}}}{\delta }_{(e,f)}^{†}{f}^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}a{f}^{{\displaystyle \frac{1}{2}}}\right]\hfill \\ \hfill =& e^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}a{e}^{{\displaystyle \frac{1}{2}}}{\delta }_{(e,f)}^{†}{f}^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}xa{e}^{{\displaystyle \frac{1}{2}}}{\delta }_{(e,f)}^{†}{f}^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}a{f}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill =& e^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}a{e}^{{\displaystyle \frac{1}{2}}}{\delta }_{(e,f)}^{†}[f^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}xa{e}^{{\displaystyle \frac{1}{2}}}+f^{-{\displaystyle \frac{1}{2}}}(1-a_{(e,f)}^{†}a)e^{{\displaystyle \frac{1}{2}}}]{\delta }_{(e,f)}^{†}{f}^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}a{f}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill =& e^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}a{e}^{{\displaystyle \frac{1}{2}}}{\delta }_{(e,f)}^{†}\delta {\delta }_{(e,f)}^{†}{f}^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}a{f}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill =& e^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}a{e}^{{\displaystyle \frac{1}{2}}}{\delta }_{(e,f)}^{†}{f}^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}a{f}^{{\displaystyle \frac{1}{2}}}=z.\hfill \end{array}$$

By hypothesis, we have

$$\left[a{e}^{{\displaystyle \frac{1}{2}}}\right]{\delta }_{(e,f)}^{†}\left[f^{-{\displaystyle \frac{1}{2}}}(1-a_{(e,f)}^{†}a)\right]=\left[(1-a_{(e,f)}^{†}a)e^{{\displaystyle \frac{1}{2}}}\right]{\delta }_{(e,f)}^{†}\left[f^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}\right]=0.$$

Then

$$\begin{array}{cc}& \delta {\delta }_{(e,f)}^{†}{f}^{-{\displaystyle \frac{1}{2}}}(1-a_{(e,f)}^{†}a)e^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill =& f^{-{\displaystyle \frac{1}{2}}}\left[(1-a_{(e,f)}^{†}a)e^{{\displaystyle \frac{1}{2}}}{\delta }_{(e,f)}^{†}\right]f^{-{\displaystyle \frac{1}{2}}}(1-a_{(e,f)}^{†}a)e^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill =& f^{-{\displaystyle \frac{1}{2}}}{e}^{{\displaystyle \frac{1}{2}}}{\delta }_{(e,f)}^{†}{f}^{-{\displaystyle \frac{1}{2}}}(1-a_{(e,f)}^{†}a)e^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill =& f^{-{\displaystyle \frac{1}{2}}}(1-a_{(e,f)}^{†}a)e^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

Thus, we have

$$\begin{array}{cc}& \gamma z\gamma -\gamma \hfill \\ \hfill =& [f^{-{\displaystyle \frac{1}{2}}}\left(a_{(e,f)}^{†}xa{e}^{{\displaystyle \frac{1}{2}}}{\delta }_{(e,f)}^{†}{f}^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}a{f}^{{\displaystyle \frac{1}{2}}}\right]\left[f^{-{\displaystyle \frac{1}{2}}}\left(a_{(e,f)}^{†}xa\right)e^{{\displaystyle \frac{1}{2}}}\right]-f^{-{\displaystyle \frac{1}{2}}}\left(a_{(e,f)}^{†}xa\right)e^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill =& \left(f^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}xa{e}^{{\displaystyle \frac{1}{2}}}\right){\delta }_{(e,f)}^{†}\left(f^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}xa{e}^{{\displaystyle \frac{1}{2}}}\right)-f^{-{\displaystyle \frac{1}{2}}}\left(a_{(e,f)}^{†}xa\right)e^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill =& \left(f^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}xa{e}^{{\displaystyle \frac{1}{2}}}\right){\delta }_{(e,f)}^{†}\left(f^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}xa{e}^{{\displaystyle \frac{1}{2}}}\right)-f^{-{\displaystyle \frac{1}{2}}}\left(a_{(e,f)}^{†}xa\right)e^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill =& [f^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}xa{e}^{{\displaystyle \frac{1}{2}}}+f^{-{\displaystyle \frac{1}{2}}}(1-a_{(e,f)}^{†}a)e^{{\displaystyle \frac{1}{2}}}]{\delta }_{(e,f)}^{†}\left(f^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}xa{e}^{{\displaystyle \frac{1}{2}}}\right)-f^{-{\displaystyle \frac{1}{2}}}\left(a_{(e,f)}^{†}xa\right)e^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill =& \delta {\delta }_{(e,f)}^{†}\left(f^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}xa{e}^{{\displaystyle \frac{1}{2}}}\right)-f^{-{\displaystyle \frac{1}{2}}}\left(a_{(e,f)}^{†}xa\right)e^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill =& \delta {\delta }_{(e,f)}^{†}[f^{-{\displaystyle \frac{1}{2}}}{a}_{(e,f)}^{†}xa{e}^{{\displaystyle \frac{1}{2}}}+f^{-{\displaystyle \frac{1}{2}}}(1-a_{(e,f)}^{†}a)e^{{\displaystyle \frac{1}{2}}}]\hfill \\ \hfill -& \delta {\delta }_{(e,f)}^{†}{f}^{-{\displaystyle \frac{1}{2}}}(1-a_{(e,f)}^{†}a)e^{{\displaystyle \frac{1}{2}}}-f^{-{\displaystyle \frac{1}{2}}}\left(a_{(e,f)}^{†}xa\right)e^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill =& \delta {\delta }_{(e,f)}^{†}\delta -\delta +f^{-{\displaystyle \frac{1}{2}}}(1-a_{(e,f)}^{†}a)e^{{\displaystyle \frac{1}{2}}}-\delta {\delta }_{(e,f)}^{†}{f}^{-{\displaystyle \frac{1}{2}}}(1-a_{(e,f)}^{†}a)e^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill =& \delta {\delta }_{(e,f)}^{†}\delta -\delta .\hfill \end{array}$$

Hence,

$$e^{-{\displaystyle \frac{1}{2}}}p{e}^{{\displaystyle \frac{1}{2}}}[\alpha z\alpha -\alpha ]=e^{-{\displaystyle \frac{1}{2}}}p{e}^{{\displaystyle \frac{1}{2}}}\beta {\beta }_{(e,f)}^{†}\beta -\beta ,$$

and then

$$\alpha z\alpha -\alpha =\beta {\beta }_{(e,f)}^{†}\beta -\beta \in R_{(e,f)}^{qnil}.$$

Since ${\left[e\beta {\beta }_{(e,f)}^{†}\right]}^{*}=e\beta {\beta }_{(e,f)}^{†},$ we have

$${\left[e(e^{-{\displaystyle \frac{1}{2}}}pxp{f}^{{\displaystyle \frac{1}{2}}}+e^{-{\displaystyle \frac{1}{2}}}(1-p)f^{{\displaystyle \frac{1}{2}}}){\beta }_{(e,f)}^{†}\right]}^{*}=e(e^{-{\displaystyle \frac{1}{2}}}pxp{f}^{{\displaystyle \frac{1}{2}}}+e^{-{\displaystyle \frac{1}{2}}}(1-p)f^{{\displaystyle \frac{1}{2}}}){\beta }_{(e,f)}^{†}.$$

Then

$${\left[e^{{\displaystyle \frac{1}{2}}}(pxp+1-p)f^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}\right]}^{*}=e^{{\displaystyle \frac{1}{2}}}(pxp+1-p)f^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}.$$

This implies that

$${\left[(pxp+1-p)f^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}{e}^{-{\displaystyle \frac{1}{2}}}\right]}^{*}=(pxp+1-p)f^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}{e}^{-{\displaystyle \frac{1}{2}}}.$$

Hence,

$${\left[\left(pxp\right)f^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}{e}^{-{\displaystyle \frac{1}{2}}}p\right]}^{*}=\left(pxp\right)f^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}{e}^{-{\displaystyle \frac{1}{2}}}p.$$

Therefore

$$\begin{array}{cc}& {\left[e\left(e^{-{\displaystyle \frac{1}{2}}}\left(pxp\right)f^{{\displaystyle \frac{1}{2}}}\right)\left(f^{-{\displaystyle \frac{1}{2}}}p{f}^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}{e}^{-{\displaystyle \frac{1}{2}}}p{e}^{{\displaystyle \frac{1}{2}}}\right)\right]}^{*}\hfill \\ \hfill =& e\left(e^{-{\displaystyle \frac{1}{2}}}\left(pxp\right)f^{{\displaystyle \frac{1}{2}}}\right)\left(f^{-{\displaystyle \frac{1}{2}}}p{f}^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}\right)e^{-{\displaystyle \frac{1}{2}}}p{e}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

That is, ${\left(e\alpha z\right)}^{*}=e\alpha z.$

Then ${\left(fz\alpha \right)}^{*}=fz\alpha .$ Accordingly, $\alpha \in R_{(e,f)}^{†}$. In this case, we have

$${\alpha }_{(e,f)}^{†}=z=f^{-{\displaystyle \frac{1}{2}}}p{f}^{{\displaystyle \frac{1}{2}}}{\beta }_{(e,f)}^{†}{e}^{-{\displaystyle \frac{1}{2}}}p{e}^{{\displaystyle \frac{1}{2}}},$$

as asserted. □

Corollary 4.5.  

Let $a\in R^{†}$ and $x\in R$. Let $\gamma =a{a}^{†}xa$ and $\delta =a{a}^{†}xa+1-a{a}^{†}$. Then $\gamma \in R^{†}$ if and only if

(1)

$\delta \in R^{†};$

(2)

${\delta }^{†}(1-a^{†}a))=1-a^{†}a,\left((1-a^{†}a){\delta }^{†}{a}^{†}\right)=0.$

In this case,

$$\begin{array}{ccc}\hfill {\delta }^{†}& =\hfill & {\alpha }^{†}+1-a{a}^{†},\hfill \\ \hfill {\gamma }^{†}& =\hfill & a^{†}a{\delta }^{†}{a}^{†}a.\hfill \end{array}$$

Proof. 

This is obvious by choosing $e=f=1$ in Theorem 4.4. □

We now present the generalized weighted Moore-Penrose inverse in terms of the ordinary generalized Moore-Penrose inverse.

Theorem 4.6.  

Let R be a *-ring and let $e,f$ be square-root elements of R. The the following are equivalent:

(1)

$a\in R_{e,f}^{†}$.

(2)

$f^{{\displaystyle \frac{1}{2}}}a{e}^{-{\displaystyle \frac{1}{2}}}\in R^{†}$

In this case,

$$a_{e,f}^{†}=f^{-{\displaystyle \frac{1}{2}}}{\left(e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\right)}^{†}{e}^{{\displaystyle \frac{1}{2}}}.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Since $a\in R_{e,f}^{†}$, there exist $x,y\in R$ such that

$$a=x+y,x^{*}ey=y{f}^{-1}{x}^{*}=0,x\in R_{(e,f)}^{†},y\in R_{(e,f)}^{qnil}.$$

By virtue of Theorem 2.7, $e^{{\displaystyle \frac{1}{2}}}x{f}^{-{\displaystyle \frac{1}{2}}}\in R^{†}$ and

$$x_{f,e}^{†}=f^{-{\displaystyle \frac{1}{2}}}{\left(e^{{\displaystyle \frac{1}{2}}}x{f}^{-{\displaystyle \frac{1}{2}}}\right)}^{†}{e}^{{\displaystyle \frac{1}{2}}}.$$

Obviously, we have

$$e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}=e^{{\displaystyle \frac{1}{2}}}x{f}^{-{\displaystyle \frac{1}{2}}}+e^{{\displaystyle \frac{1}{2}}}y{f}^{-{\displaystyle \frac{1}{2}}}.$$

We directly verify that

$$\left[e^{{\displaystyle \frac{1}{2}}}y{f}^{-{\displaystyle \frac{1}{2}}}\right]{\left[e^{{\displaystyle \frac{1}{2}}}y{f}^{-{\displaystyle \frac{1}{2}}}\right]}^{*}=e^{{\displaystyle \frac{1}{2}}}y{f}^{-1}{y}^{*}{e}^{{\displaystyle \frac{1}{2}}}=e^{{\displaystyle \frac{1}{2}}}\left(y{f}^{-1}{y}^{*}e\right))e^{-{\displaystyle \frac{1}{2}}}\in R^{qnil}.$$

Furthermore, we check that

$$\begin{array}{ccc}\hfill {\left(e^{{\displaystyle \frac{1}{2}}}x{f}^{-{\displaystyle \frac{1}{2}}}\right)}^{*}\left(e^{{\displaystyle \frac{1}{2}}}y{f}^{-{\displaystyle \frac{1}{2}}}\right)& =\hfill & f^{-{\displaystyle \frac{1}{2}}}\left[x^{*}ef\right]f^{-{\displaystyle \frac{1}{2}}}]=0,\hfill \\ \hfill \left(e^{{\displaystyle \frac{1}{2}}}y{f}^{-{\displaystyle \frac{1}{2}}}\right){\left(e^{{\displaystyle \frac{1}{2}}}x{f}^{-{\displaystyle \frac{1}{2}}}\right)}^{*}& =\hfill & e^{{\displaystyle \frac{1}{2}}}\left[y{f}^{-1}{x}^{*}\right]e^{{\displaystyle \frac{1}{2}}}=0.\hfill \end{array}$$

By virtue of Corollary ???, we have $e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\in R^{†},$ as required.

$\left(2\right)\Rightarrow \left(1\right)$ Set $x={\left(e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\right)}^{†}.$ We claim that $a_{e,f}^{†}=f^{-{\displaystyle \frac{1}{2}}}x{e}^{{\displaystyle \frac{1}{2}}}.$ Then there exist $x,y\in R$ such that

$$e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}=x+y,x^{*}y=y{x}^{*}=0,x\in R^{†},y{y}^{*}\in R^{qnil}.$$

Then we have

$$a=e^{-{\displaystyle \frac{1}{2}}}x{f}^{{\displaystyle \frac{1}{2}}}+e^{-{\displaystyle \frac{1}{2}}}y{f}^{{\displaystyle \frac{1}{2}}}.$$

Claim 1. $e^{-{\displaystyle \frac{1}{2}}}x{f}^{{\displaystyle \frac{1}{2}}}\in R_{e,f}^{†}.$ Since $e,f\in R$ are square-root elements, so are $e^{-1}$ and $f^{-1}$. Applying Theorem 2.7 to $e^{-1}$ and $f^{-1}$, we see that $e^{-{\displaystyle \frac{1}{2}}}x{f}^{{\displaystyle \frac{1}{2}}}\in R_{e,f}^{†}.$

Claim 2. $e^{-{\displaystyle \frac{1}{2}}}y{f}^{{\displaystyle \frac{1}{2}}}\in R_{e,f}^{qnil}$. We verify that

$$\left(e^{-{\displaystyle \frac{1}{2}}}y{f}^{{\displaystyle \frac{1}{2}}}\right)f^{-1}{\left(e^{-{\displaystyle \frac{1}{2}}}y{f}^{{\displaystyle \frac{1}{2}}}\right)}^{*}e=e^{-{\displaystyle \frac{1}{2}}}y{y}^{*}{e}^{{\displaystyle \frac{1}{2}}}\in R^{qnil}.$$

Obviously, we verify that ${\left(e^{-{\displaystyle \frac{1}{2}}}x{f}^{{\displaystyle \frac{1}{2}}}\right)}^{*}e\left(e^{-{\displaystyle \frac{1}{2}}}y{f}^{{\displaystyle \frac{1}{2}}}\right)={\left(f^{{\displaystyle \frac{1}{2}}}\right)}^{*}\left(x^{*}y\right)f^{{\displaystyle \frac{1}{2}}}=0$ and $\left(e^{-{\displaystyle \frac{1}{2}}}y{f}^{{\displaystyle \frac{1}{2}}}\right)f^{-1}{\left(e^{-{\displaystyle \frac{1}{2}}}x{f}^{{\displaystyle \frac{1}{2}}}\right)}^{*}=e^{-{\displaystyle \frac{1}{2}}}\left(y{x}^{*}\right){\left(e^{-{\displaystyle \frac{1}{2}}}\right)}^{*}=0$. This completes the proof. □

Corollary 4.7.  

Let $\mathcal{A}$ be a unital $C^{*}$-algebra and let $e,f$ be positive invertible elements of $\mathcal{A}$. The the following are equivalent:

(1)

$a\in {\mathcal{A}}_{e,f}^{†}$.

(2)

$f^{{\displaystyle \frac{1}{2}}}a{e}^{-{\displaystyle \frac{1}{2}}}\in {\mathcal{A}}^{†}$

In this case,

$$a_{e,f}^{†}=f^{-{\displaystyle \frac{1}{2}}}{\left(e^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}\right)}^{†}{e}^{{\displaystyle \frac{1}{2}}}.$$

Proof. 

Since e and f are positive, they are square-root elements in $\mathcal{A}$. This completes the proof by Theorem 4.6. □

Corollary 4.8.  

Let $\mathcal{A}$ be a ring and let $e,f,h$ be square-root elements of $\mathcal{A}$. Then the following are equivalent:

(1)

${\left(ab\right)}_{e,h}^{†}=b_{e,f}^{†}{a}_{f,h}^{†}$.

(2)

${\left(a_1{b}_1\right)}^{†}=b_1^{†}{a}_1^{†}$,

where $a_1=h^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}$ and $b_1=f^{{\displaystyle \frac{1}{2}}}b{e}^{-{\displaystyle \frac{1}{2}}}$.

Proof. 

Let $a_1=h^{{\displaystyle \frac{1}{2}}}a{f}^{-{\displaystyle \frac{1}{2}}}$ and $b_1=f^{{\displaystyle \frac{1}{2}}}b{e}^{-{\displaystyle \frac{1}{2}}}$. Then $a_1{b}_1=h^{{\displaystyle \frac{1}{2}}}\left(ab\right)e^{-{\displaystyle \frac{1}{2}}}$. In view of Theorem 4.6, we see that $a\in {\mathcal{A}}_{f,h}^{†},b\in {\mathcal{A}}_{e,f}^{†},ab\in {\mathcal{A}}_{e,h}^{†}$ if and only if $a_1,b_1,a_1{b}_1\in {\mathcal{A}}^{†}$. Suppose the preceding weighted MP inverses exist. In view of Theorem 4.6, we have

$$\begin{array}{ccc}\hfill a_{f,h}^{†}& =\hfill & f^{-{\displaystyle \frac{1}{2}}}{a}_1^{†}{h}^{{\displaystyle \frac{1}{2}}},\hfill \\ \hfill b_{e,f}^{†}& =\hfill & e^{-{\displaystyle \frac{1}{2}}}{b}_1^{†}{f}^{{\displaystyle \frac{1}{2}}},\hfill \\ \hfill {\left(ab\right)}_{e,h}^{†}& =\hfill & e^{-{\displaystyle \frac{1}{2}}}{\left(h^{{\displaystyle \frac{1}{2}}}\left(ab\right)e^{-{\displaystyle \frac{1}{2}}}\right)}^{†}{h}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

Therefore ${\left(ab\right)}_{e,h}^{†}=b_{e,f}^{†}{a}_{f,h}^{†}$ if and only if

$$e^{-{\displaystyle \frac{1}{2}}}{\left(a_1{b}_1\right)}^{†}{h}^{{\displaystyle \frac{1}{2}}}=e^{-{\displaystyle \frac{1}{2}}}{b}_1^{†}{a}_1^{†}{h}^{{\displaystyle \frac{1}{2}}}$$

if and only if ${\left(a_1{b}_1\right)}^{†}=b_1^{†}{a}_1^{†},$ as asserted. □